I\'ve been working on this problem with my friend for awhile nowand we are havin
ID: 684891 • Letter: I
Question
I've been working on this problem with my friend for awhile nowand we are having some issues. We believe that the formula isC6H8O based on the spectra provided but weare having trouble identifying the compound. We are using AceOrganic Online Homework and we have 10 tries at eachquestion. Any help would be appreciated!
You will need to measure the heights of the integral lines toestimate the area under each signal. Pay attention tomultiplicities when they are visible. When given a molecularformula, always calculate degrees of unsaturation before beginningto put together the molecular puzzle. Note that you are limited to10 attempts for each problem, so work out your answers first onpaper. (19) The following spectra are taken from acompound that is an important starting material for organicsynthesis. Determine its structure considering all the spectratogether. Fig. 1 Click image to enlarge){kind=link}
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I've been working on this problem with my friend for awhile nowand we are having some issues. We believe that the formula isC6H8O based on the spectra provided but weare having trouble identifying the compound. We are using AceOrganic Online Homework and we have 10 tries at eachquestion. Any help would be appreciated!
Explanation / Answer
Firstly, I agree that the formula is C6H8O. Igot 2-cyclohexen-1-one for the structure. If you don't want to readthrough my long-winded explanation (I sorta wrote this as I went),skip down to the Summary! =P The MS shows an M-peak of 96 - I was taught something called the'rule of 13', which basically tells you to divide the weight by 13(= mass of CH2), and add the left-over as hydrogens: 96 / 13 = 7 + 5 = C7H7 +5H = C7H12However, the IR shows a strongpeak at 1685 cm-1, which suggests the presence of, forexample, an aryl-conjugated C=C, a C=C-O, aryl or unsaturatedaldehydes/ketones, aryl or unsaturated acids... etc. (Butnot saturated aldehydes or ketones, which show up at over1700 cm-1.) The H NMR spectrum also shows some lowfield(= high ppm, high frequency) peaks that aren't aromatic, which alsosuggests the presence of an electronegative group. Let's assume we have one oxygen. The MW of oxygen is 16, so modifythe chemical formula by subtracting 1*C + 4*H (=12+4=16) to get thenew formula: C6H8O, which is what youhad. This means we have 6 - 8/2 + 1 = 3 double bond equivalents (DBE=number of double bonds and/or rings). I tend to head for the H NMR spectra first. The bunch of peaks at 7ppm would usually make me think aromatic, but we don'thave enough DBEs for that (aromatic = 4 DBEs). The second thing Inoticed is that the 2.2 and 2.5 ppm have also got prettycomplicated splitting, which makes me think cyclic, sincethe spectra of cyclic molecules is often complicated due tocis/trans stereochemistry effects. The most common cyclic molecule used is, in my opinion,cyclohexane. The coupling constant for a hydrogen atom that iscis to a hydrogen on an adjacent carbon in 8-13 Hz, and2-6 Hz for trans, which is sort of close to the 7 Hz inthe spectrum. Looking at the C NMR spectrum, we notice that thereare no quaternary (q) peaks so we have no CH3 'ends' onour molecule, which also supports a cyclic structure. Assuming this is correct, this means we have no more carbon atomsleft to use, and we only need to be concerned with H's and the O,and we have 2 more DBE's left. We'll deal with the oxygen first. We know that there can't be anyacids (since we don't have enough O's) or alcohols (IR spectrumshows no alcohol peak, and there are no singlet peaks in the HNMR). We can't have an aldehyde either, since we have a cyclicmolecule. This means the oxygen must be a ketone coming off thering = cyclohexanone. Going back to the C NMR spectrum, note thatthe C=O carbon must correspond to the 200 ppm singlet (= nodirectly attached H's) peak, which is appropriate for a ketone. We now have only one DBE left to deal with, and putting this inwill cut down the H's to the correct amount. So where do we put it?There are only two choices (due to symmetry): if I label the ketonecarbon '1', then the double bond can either go between carbons 2and 3, or 3 and 4. However, if we put it between C3 and C4 the H NMR wouldnot show any doublets - and we know we have a doublet at 6ppm. SUMMARY The structure is 2-cyclohexen-1-one: H NMR: The 7.0-7.1 ppm (1H) pentet corresponds to the C3 hydrogen, whichis split by the C2 hydrogen, the two C4 hydrogens, and maybesplitting from one of the hydrogens on C5 (though I'm not quitesure about that). The 6.0-6.1 ppm (1H) doublet is from the C2 hydrogen, which issplit with a 7 Hz constant by the C3 hydrogen. The 2.5 ppm (4H) multiplet probably corresponds to the hydrogens onC5 and C6 (?) The 2.0-2.1 ppm (2H) multiplet probably corresponds to the C4hydrogens, which are farthest away from the ketone and thereforeresonate at the lowest ppm. They are shielded because of theresonance/tautomerism of the ketone (electrons move from the ketonedouble bond to the C1-C2 bond, and the C2-C3 double bond moves toC3-C4, meaning C4 has extra electron density, and the C4 hydrogensare shielded). C NMR: The 200 ppm singlet corresponds to C1 = ketone carbon with noattached hydrogens. The 150 ppm doublet corresponds to C3, which has 1 attachedhydrogen, because: The only other doublet, at 130 ppm, correspondsto C2. C2 has extra shielding due to the resonance mentionedearlier, and is therefore lower ppm than C3. The 38 ppm triplet corresponds to C6, which carries two hydrogens,and is closest to the ketone, and is therefore deshielded. The 22 and 25 ppm triplets correspond to C4 and C5. I think C5would have the lower value because it's furthest away fromboth the ring and the ketone double bonds, while C4(opposite the ketone) is adjacent to the ring double bond. IR: 1685 cm-1 corresponds to the ketone. No other featuresof note, other than the fact that the absence of some signals cantell us about functionalities that are not in our molecule(e.g. alcohols). MS: The M=96 peak corresponds to the molecular weight (MW) ofC6H8O. I think the 68 peak corresponds to either ring opening, followed byloss of the C(6)H2 and C(5)H2 groups (MW = 28= 96-68), or to ring opening followed by loss of the oxygen andC(1) (MW = 16 + 12 = 28).
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